Double integrals

[Tolaso J Kos]

Evaluate the following integrals:

\( \rm{i.} \) \( \displaystyle \int_{0}^{1}\int_{0}^{1}\frac{x-y}{\left ( x+y \right )^3}\,dy\,dx \)

\( \rm{ii.} \) \( \displaystyle \int_{0}^{1}\int_{0}^{1}\frac{x-y}{\left ( x+y \right )^3}\,dx\,dy \)

[Papapetros Vaggelis]

Hello Tolis.

\(\displaystyle{\rm{i.}}\) If \(\displaystyle{x\in\left[0,1\right]}\) , then :

\(\displaystyle{\begin{aligned} \int_{0}^{1}\dfrac{x-y}{\left(x+y\right)^3}\,\mathrm{d}y&=\int_{0}^{1}\left(x-y\right)\,d\,\left(-\dfrac{1}{2}\,\dfrac{1}{\left(y+x\right)^2}\right)\\&=\left[\dfrac{y-x}{2\,\left(y+x\right)^2}\right]_{0}^{1}-\int_{0}^{1}-\dfrac{1}{2}\,\dfrac{1}{\left(y+x\right)^2}\,d\,\left(x-y\right)\\&=\dfrac{1-x}{2\,\left(1+x\right)^2}+\dfrac{1}{2\,x}-\dfrac{1}{2}\,\int_{0}^{1}\dfrac{1}{\left(y+x\right)^2}\,\mathrm{d}y\\&=\dfrac{1}{2\,\left(1+x\right)}-\dfrac{x}{\left(1+x\right)^2}+\dfrac{1}{2\,x}+\left[\dfrac{1}{2\,\left(y+x\right)}\right]_{0}^{1}\\&=\dfrac{1}{2\,\left(1+x\right)}-\dfrac{x}{\left(1+x\right)^2}+\dfrac{1}{2\,x}+\dfrac{1}{2\,\left(x+1\right)}-\dfrac{1}{2\,x}\\&=\dfrac{1}{x+1}-\dfrac{x}{\left(1+x\right)^2}\\&=\dfrac{1}{\left(1+x\right)^2}\end{aligned}}\)

so :

\(\displaystyle{\begin{aligned}I=\int_{0}^{1}\,\int_{0}^{1}\dfrac{x-y}{\left(x+y\right)^3}\,\mathrm{d}y\,\mathrm{d}\,x&=\int_{0}^{1}\dfrac{1}{\left(1+x\right)^2}\,\mathrm{d}x\\&=\left[-\dfrac{1}{1+x}\right]_{0}^{1}\\&=\dfrac{1}{2}\end{aligned}}\) .

\(\displaystyle{\rm{ii.}}\) We have that :

\(\displaystyle{\int_{0}^{1}\,\int_{0}^{1}\dfrac{x-y}{\left(x+y\right)^3}\,\mathrm{d}x\,\mathrm{d}y=-\int_{0}^{1}\,\int_{0}^{1}\,\dfrac{y-x}{\left(y+x\right)^3}\,\mathrm{d}x\,\mathrm{d}y=-I=-\dfrac{1}{2}}\) .